Problem: Let $y=e^{(3x^2-4)}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $ \ln(x)\cdot e^{(3x^2-4)}$ (Choice B) B $ e^{(3x^2-4)}$ (Choice C) C $ e^{(6x)}$ (Choice D) D $6x\cdot e^{(3x^2-4)}$
Answer: $e^{(3x^2-4)}$ is an exponential function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=3x^2-4$, then $y=e^{u(x)}$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[e^{u(x)}\right]=e^{u(x)}\cdot u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}e^{(3x^2-4)} \\\\ &=\dfrac{d}{dx}e^{u(x)}&&\gray{\text{Let }u(x)=3x^2-4} \\\\ &=e^{u(x)}\cdot u'(x) \\\\ &=e^{(3x^2-4)}\cdot (6x)&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=6x\cdot e^{(3x^2-4)} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=6x\cdot e^{(3x^2-4)}$.